What is the solution for y^{''}+2y^'+y=4e^x ?

The solution for y^{''}+2y^'+y=4e^x is y=c_{1}e^{-x}+c_{2}xe^{-x}+e^x

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y^{''}+2y^'+y=4e^x

y^{^{''}} + 2 y^{^{'}} + y = 4 e^{x}

y = c_{1} e^{- x} + c_{2} x e^{- x} + e^{x}

Solution steps

y^{''} (x) + 2 y^{'} (x) + y = 4 e^{x}

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} + e^{x}

y = c_{1} e^{- x} + c_{2} x e^{- x} + e^{x}

y^{^{''}} - 2 y^{^{'}} + y = e^{t - 1}, y (1) = 0, y^{^{'}} (1) = 5

y^{^{''}} - y^{^{'}} = x e^{2 x}

y^{^{''}} + 4 y = - 8 sin (2 t)

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2} x (cos (x) - 9 sin (x))

\frac{d ^{2} y}{d t ^{2}} - 2 \frac{d y}{d t} + y = 5 e^{t}

What is the solution for y^{''}+2y^'+y=4e^x ?
The solution for y^{''}+2y^'+y=4e^x is y=c_{1}e^{-x}+c_{2}xe^{-x}+e^x